# Driving A Car Straight Up

A colleague and I were playing around with a videogame physics engine and we started speculating about whether a particular driving stunt would be doable in the real world.

Imagine a ramp that curves up and ends with the top pointing straight up. You get in a muscle car, aim it at the ramp, and floor it. By the time you hit the ramp you’re travelling, say, 110 or 120 mph. Assume a gentle enough curve to the ramp that you don’t bottom out.

1. How high could you go? (Some rough back-of-the-envelope calculations we did suggested 100m might be possible.)

2. Would you have enough time at the peak of your trajectory to bail out and parachute to the ground?

3. Is there any way this stunt could be survivable without bailing out of the car?

4. Wouldn’t this be cool as hell to watch?

I can’t do all the calculations now but it would be the same as using a ramp to launch a car straight upwards. Once the car leaves the ramp and heads directly up, it will start to decelerate as gravity pulls it back down.
Even it there is “track” on the vertical portion of your ramp it makes no difference since your car will have no traction to further accelerate or maintain speed at that point.

That’s what I got, assuming kinetic energy (of the 100mph car) is converted to potential energy without loss. In reality there’ll be some inefficiencies, but the car can also apply power while on the ramp.

If the car climbs the ramp, then roll back down with all wheels in contact with the ramp, I think it should be survivable.

Don’t forget, you’re assuming a ‘perfect’ car in an ideal enviroment. I would think in real life once you’re pointed stright up, not only are you going to lose traction with the ramp, but I would guess the weight of the engine/cargo (ie driver’ and anything on the top half (the half of the car away from the ramp) will cause it to lean back and land on the roof.

No, no, I want it to be airborne … .

You could probably overcome some of the physics with a special car. I envision a badass Barracuda with wings and a rudder, a Jet engine for initial thrust and two rocket boosters timed to trip at gravitas.

there are race cars (the mclaren f1 is one example) that generate enough downforce at speed to theoretically drive upside down and still maintain traction due to aerodynamics.

Shouldn’t that be in SPOILER tags?

At 120 mph, your energy is the same as it would be about 480’ above that with no speed. This ignores frictional losses, so let’s assume that you might gain as much as 400’

At 400’ with zero velocity, your chance of a successful bailout would be low (though certainly not zero). Might be a good idea to include an ejection system.

Sure: rig a net to catch the car; fit the car with a parachute; use giant airbags. Plenty of others.

And they can do that at significantly less than their maximum speed, so I’d guess that even fighting gravity, those cars could drive straight up.

Until the air started to get thinner, that is.

Assuming that you aimed perfectly, and caught the car (after it was airborne) with a similar-shaped ramp, the landing could in principle be as gentle as the launch. In practice, I suspect it would be difficult to land so precisely.

If I’m not mistaken, the key modification you must make to your test vheicle to ensure success would be to replace the stock horn with one that plays* Dixie*, set to “honk” as soon as you are airbone.

They’d run out of speed.
Even if they generated 1500 ft-lb of torque, they’d wind up, at some point, losing their accumulated speed and moving upwards at a rate of speed of:

[weight in lbs] / [ ft-lb of peak torque ] feet per second

so, if it was a 3000 lb race car, 2 feet per second or 7200 feet per hour. That’s right around 1.4 MPH, so they’d probably run out of downforce. This is assuming that your race car has a stump-puller first gear, which I’ve never heard in that category of vehicle.

By, “Due to aerodynamics”, do you mean that the car is designed so that wind pushes it down (toward the road)? Just looking for clarification.

Yes. That’s what those big “wings” on the rear of the car are for. They are basically upside-down airfoils, that push the car down rather than lift it up. Some cars have wings on the front end as well, and the underside of the car can be shaped to suck it down even further. The F1 regulations limit the amount of aerodynamic downforce a car can have to try and even the field somewhat. (The regs try to balance the effect of car engineering and driver skills to make both good car design and good driving necessary to win).

Not really the same thing, but sailplane pilots (a lot of them anyway) regularly do something similar. Known as a race finish. It is common for the pilot to cross the finish line at quite low altitude…50’ AGL would be high. Excess airspeed is then converted to altitude with a ~2g pullup into a ~0g climb(the “zoom”) , then a ~-`1/2g pushover to a normal landing pattern. 120kt at ground level will put you at ~1000’AGL 50-60kt. for your landing pattern. Altitude gain is maximised by a near vertical “zoom” phase. At zero-g the wing is producing no lift, so induced drag is minimal. The tradeoff is in the pullup phase. The harder the pull, the more energy lost to induced drag. A 2g pull is near optimal for most sailplanes.

If the pilot were to hold zero-g untill airspeed dropped to zero, the resulting tailslide can be fairly violent as the gliders airdynamics turn it around. Since most cars are not very stable in reverse, your scenerio would suffer similar problems unless the car were specially designed for stability while in reverse.

Anyway, if you hire an aerobatic sailplane ride, the pilot will be happy to demonstrate so you can experience this first hand. Possibly even a “normal” sightseeing ride could include such, but if it is specifically an aerobatic ride, then you are gauranteed that both the aircraft and pilot will be up to the task.

Also, I think there up-and-back roller coasters based on the same thing.

So the cool thing would be to build a properly sloping ramp on the side of a skyscraper, drive the F1 car straight up the side, and then bail out w/ a parachute once you run out of wall and start to reach the top of your climb.

I nominate Pochacco to take the first drive.

Are you sure this is correct?
First, F1 cars only weigh about 550 kgs according to this page. Even with driver, that’s still half of what you estimated. For this calculation, let’s assume a 700 kg car + driver.

F1 cars accelerate faster than 1g when on the ground. This site claims that they do 0-100mph in 3.6 seconds, which at constant acceleration is approx 12.5 m/s/s. The acceleration is not going to be constant; it’s going to vary with the normal force. But, just estimating with constant acceleration, going straight up, the cars should accelerate until air resistance plus gravity provides for -12.5 m/s/s acceleration.

Gravity provides for 9.8 m/s/s, so we need drag to provide an additional… lets say 3 m/s/s of deceleration.

According to this site

F[sub]d[/sub] = 1/2 * p * C[sub]D[/sub] * A * v[sup]2[/sup]. According to Wikipedia, the coefficient of drag for F1 cars is “close to 1”. The density of air is 1.2 kg/m[sup]3[/sup] at STP. Randomly estimating the frontal area of an F1 car as about 2 m[sup]2[/sup], and solving for v, we get

v = sqrt( 2 * F[sub]d[/sub] / (p * C[sub]D[/sub] * A))
v = sqrt( 2 * 2100 N / (1.2 kg/m[sup]3[/sup] * 2 m[sup]3[/sup])
v approx = 40 m/s or 90 mph.

So, assuming all that stuff I did (and assuming I didn’t make any mistakes in that calculation), it looks like the steady state of the car would be around 90 mph straight up. Would that be enough to maintain sufficient downforce to get the acceleration required? I have no idea.

However, it should be pointed out that F1 cars have an absurdly high coefficient of drag compared to what they could. They need it to create all the downforce that enables them to go through tight turns. Even if an F1 car couldn’t do it, a more aerodynamic car might be able to.

This already exists as an amusement park ride, although it’s on rails:

http://www.sixflags.com/parks/magicmountain/Rides/superman.html

I’ve ridden it several times, lots of fun. It’s 0-100mph in about 4 seconds (they say 7 seconds but I think that’s the time to reach maximum altitude, at which point you are going 0mph and begin the backslide). There’s no power or pulling mechanism after the straightaway, you just coast up.

In theory you could bail out at the top - BASE jumpers certainly jump from that height. Ejection seats work at zero airspeed and zero velocity.

From a practical standpoint you’d need to be a careful driver on the way back down to avoid any kind of skid or roll, that would be disastrous.